What is the extraneous solution to these equations? $\dfrac{x^2 - 2}{x - 1} = \dfrac{-7x + 6}{x - 1}$
Solution: Multiply both sides by $x - 1$ $ \dfrac{x^2 - 2}{x - 1} (x - 1) = \dfrac{-7x + 6}{x - 1} (x - 1)$ $ x^2 - 2 = -7x + 6$ Subtract $-7x + 6$ from both sides: $ x^2 - 2 - (-7x + 6) = -7x + 6 - (-7x + 6)$ $ x^2 - 2 + 7x - 6 = 0$ $ x^2 - 8 + 7x = 0$ Factor the expression: $ (x - 1)(x + 8) = 0$ Therefore $x = 1$ or $x = -8$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.